题目链接：https://ac.nowcoder.com/acm/contest/70/B
题目大意：
略
分析：
先DFS求出所有幸运数，然后暴力即可
代码如下：
1 #pragma GCC optimize("Ofast") 2 #include <bits/stdc++.h> 3 using namespace std; 4 5 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0); 6 #define Rep(i,n) for (int i = 0; i < (n); ++i) 7 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 8 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 9 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 10 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 11 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 12 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 13 14 #define pr(x) cout << #x << " = " << x << " " 15 #define prln(x) cout << #x << " = " << x << endl 16 17 #define LOWBIT(x) ((x)&(-x)) 18 19 #define ALL(x) x.begin(),x.end() 20 #define INS(x) inserter(x,x.begin()) 21 22 #define ms0(a) memset(a,0,sizeof(a)) 23 #define msI(a) memset(a,inf,sizeof(a)) 24 #define msM(a) memset(a,-1,sizeof(a)) 25 26 #define pii pair<int,int> 27 #define piii pair<pair<int,int>,int> 28 #define MP make_pair 29 #define PB push_back 30 #define ft first 31 #define sd second 32 33 template<typename T1, typename T2> 34 istream &operator>>(istream &in, pair<T1, T2> &p) { 35 in >> p.first >> p.second; 36 return in; 37 } 38 39 template<typename T> 40 istream &operator>>(istream &in, vector<T> &v) { 41 for (auto &x: v) 42 in >> x; 43 return in; 44 } 45 46 template<typename T1, typename T2> 47 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 48 out << "[" << p.first << ", " << p.second << "]" << " "; 49 return out; 50 } 51 52 typedef long long LL; 53 typedef unsigned long long uLL; 54 typedef pair< double, double > PDD; 55 typedef set< int > SI; 56 typedef vector< int > VI; 57 const double EPS = 1e-10; 58 const int inf = 1e9 + 9; 59 const LL mod = 1e9 + 7; 60 const int maxN = 1e5 + 7; 61 const LL ONE = 1; 62 63 LL l, r, ans; 64 LL lucky[1500], n; 65 66 // 求所有幸运数 67 inline void dfs(LL x, int cnt) { 68 if(cnt > 9) return; 69 lucky[n++] = x; 70 dfs(x*10 + 4, cnt + 1); 71 dfs(x*10 + 7, cnt + 1); 72 } 73 74 int main(){ 75 INIT(); 76 cin >> l >> r; 77 lucky[n++] = 4444444444; 78 dfs(0, 0); 79 sort(lucky, lucky + n); 80 81 int j = lower_bound(lucky, lucky + n, l) - lucky; 82 83 while(l <= r) { 84 ans += (min(lucky[j], r) - l + 1) * lucky[j]; 85 l = lucky[j++] + 1; 86 } 87 88 cout << ans << endl; 89 return 0; 90 } 91 92 /* 93 1 1000000000 94 1394672350065645019 95 96 1 1000 97 1397683 98 99 447 447477474 100 168389348342066109 101 102 1 436278568 103 163403864955643707 104 105 1 4328955 106 16190029435407 107 108 4328956 4444444 109 513284393116 110 111 4328956 436278568 112 163387674926208300 113 114 2354 543262 115 231508617956 116 117 999999999 1000000000 118 8888888888 119 */